⇔ \(-1\le-2+\dfrac{1}{m}\le1\)
⇔ \(1\le\dfrac{1}{m}\le3\)
⇔ \(\dfrac{1}{3}\le m\le1\)
\(-1\le\dfrac{-2m+1}{m}\le1\) \(\left(m\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1\le\dfrac{-2m+1}{m}\\\dfrac{-2m+1}{m}\le1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-2m+1+m}{m}\ge0\\\dfrac{-2m+1-m}{m}\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-m+1}{m}\ge0\\\dfrac{-3m+1}{m}\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0< m\le1\\\left[{}\begin{matrix}m\ge\dfrac{1}{3}\\m< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{3}\le m\le1\)
